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Probability of Multiple Failures

Date: 06/20/2002 at 21:56:14
From: Marzineh Farvon
Subject: Complete Probability in Calculus

If a car has 2 headlights with an average lifespan of 2500 hours,
whose probability of failure can be modeled by:

           f(t) = 0 if t < 0 ;
           f(t) = 1/u e ^(-t/u) if t >= 0;

for the mean value u = 2500, what is the probability that both lamps 
will fail within 2500 hours, and what is the probability that both 
will fail in a total of 2500 hours?


Date: 06/21/2002 at 07:27:50
From: Doctor Anthony
Subject: Re: Complete Probability in Calculus

The probability that either will fail within 2500 hours is

                      2500
     probability = INT     [(1/u)e^(-t/u).dt]
                      0

                 = (-u/u)e^(-t/u)  from 0 to 2500

                 = -[e^(-1) - e^0]

                 = 1 - e^(-1)

                 =  0.63212

Assuming that multiple failures would be independent events, the
probability that BOTH fail within 2500 hours

    probability =  0.63212^2 

                =  0.399576


To determine the probability that both will fail within a total of
2500 hours requires a little bookwork. 

Using 'a' as the parameter the probability distribution function (PDF)
of the exponential distribution is

  f(x) = ae^(-ax)    

and similarly 

  g(y) = ae^(-ay)

Let 

  z = x + y  

so

  y = z - x

Let S(z) be the cumulative distribution function (CDF) of z = x + y.

  S(z) = Prob(Z < z) = Prob(X+Y < z) 

                     = INT [INT [f(x) g(y) dx dy]]

                     = INT [f(x)dx INT [g(y) dy]]

and differentiating S(z) with respect to z we obtain

  S'(z) =  s(z) = INT [f(x) g(z-x) dx]

              z 
    s(z) = INT  [dx ae^(-ax) ae^(-a(z-x))]
              0 

                   z
         =  a^2 INT  [dx e^(-ax) e^(-az) e^(ax)]
                   0

                   z 
         =  a^2 INT  [dx e^(-az)]    (Note: z is constant)
                   0

         = a^2 e^(-az) z

The PDF of z is 

  s(z) = a^2 z e^(-az)   z >= 0

For this problem we have a = 1/2500 and we require

           2500
    a^2.INT     [z e^(-az) dz]      
           0

and integrating by parts, 

  = a^2 [z (-e^(-az)/a) + (1/a) INT [e^(-az) dz]]

  = a [-z e^(-az) - (1/a)e^(-az)]   from 0 to 1/a

  = a[-(1/a)e^(-1) - (1/a)e^(-1) + (1/a)e^(0)]

  =  -e^(-1) - e^-1 + 1

  =  1 - 2.e^(-1)

  = 0.264241

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Probability

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